cazinskup3

2022-06-21

Solving a system of three linear equations with three unknowns
Consider the following system of equations
$2x+2y+z=2$
$-x+2y-z=-5$
$x-3y+2z=8$
Form an augmented matrix, then reduce this matrix to reduced row echelon form and solve the system.
Given:
$2x+2y+z=2$
$-x+2y-z=-5$
$x-3y+2z=8$
Matrix form:
$\left(\begin{array}{cccc}2& 2& 1& 2\\ -1& 2& -1& -5\\ 1& -3& 2& 8\end{array}\right)$
$\left(\begin{array}{cccc}2& 0& 0& 2\\ 0& 3& 0& -3\\ 0& 0& \frac{5}{6}& \frac{5}{3}\end{array}\right)$
Solution: $x=1;y=-1;z=2;$

### Answer & Explanation

Kaydence Washington

You're hardly completely wrong! The process you describe is "spot on", and yes, your solution is correct.
You could row reduce a bit further, but there was really no need here.
You've successfully solved the system of equations.

vittorecostao1

You can reduce your matrix further. remember that you can multiply and/or divide each row so you end up obtaining
$\left(\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& -1\\ 0& 0& 1& 2\end{array}\right)$

Do you have a similar question?

Recalculate according to your conditions!